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Math Help - Conditional Propositions & Logical Equivalence

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    Conditional Propositions & Logical Equivalence

    I know this is probably really easy, but I am having the hardest time understanding what I am supposed to be doing and how I am supposed to be doing it.

    It says, for each pair of propositions P & Q, state whether P & Q are logically equivalent.

    P= p^q, Q= ~p v~q

    Ok, so i don't know really where to start, I know I draw a truth table and have p q on the left side and I have TT, TF, FT, FF....then on the right side I don't know what to do.

    Any help, please?
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    Quote Originally Posted by dude15129 View Post
    I know this is probably really easy, but I am having the hardest time understanding what I am supposed to be doing and how I am supposed to be doing it.

    It says, for each pair of propositions P & Q, state whether P & Q are logically equivalent.

    P= p^q, Q= ~p v~q

    Ok, so i don't know really where to start, I know I draw a truth table and have p q on the left side and I have TT, TF, FT, FF....then on the right side I don't know what to do.

    Any help, please?
    your table should look like this:

    \begin{array}{|c|c|c|c|c|c|}<br />
\hline p & q & \neg p & \neg q & p \wedge q & \neg p \vee \neg q \\<br />
\hline T & T & F & F & & \\<br />
\hline T & F & F & T & & \\<br />
\hline F & T & T & F & & \\<br />
\hline F & F & T & T & & \\<br />
\hline <br />
           \end{array}

    now lets start with the basic, do you know what ^ and v mean?
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    yes, the ^ is conjunction (and) and the v is disjunction (or).

    And I have that the tables for p^q are T,F,F,F and p v q are T,T,T,F.

    but I don't know really what i'm doing with those.
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    Quote Originally Posted by dude15129 View Post
    yes, the ^ is conjunction (and) and the v is disjunction (or).

    And I have that the tables for p^q are T,F,F,F and p v q are T,T,T,F.

    but I don't know really what i'm doing with those.
    shouldn't you be looking at ~p v ~q ?

    the thing is, they are equivalent if they have the same truth value in each instance. so they must both be true or both be false at the same time. otherwise, they are not equivalent
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    ok, so for ~p v ~q, the statement will be true if at least one of them is false. So it would go as follows: T,T,T,F.

    And for the statement p^q, it is: T,F,F,F.

    And since the two are not the same, P is not equal to Q.
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    Quote Originally Posted by dude15129 View Post
    ok, so for ~p v ~q, the statement will be true if at least one of them is false. So it would go as follows: T,T,T,F.

    And for the statement p^q, it is: T,F,F,F.

    And since the two are not the same, P is not equal to Q.
    actually ~p v ~q would produce the following: F, T, T, T

    This is because
    ~p v ~q is a negation of p ^ q

    ie. ~(
    p ^ q) = ~p v ~q which is known as De Morgan's Law.
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    ok, I see it now...and for the final answer you would get P is not equal to Q. right?

    I have another question, too.
    Another problem says, P= p ^ (q v r), Q= (p v q) ^ (p v r)
    so does the table have p, q, r columns to start off with, leaving 8 possible T/F combinations?
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    Quote Originally Posted by dude15129 View Post
    ok, I see it now...and for the final answer you would get P is not equal to Q. right?
    not equivalent, yes. in fact, Q = ~P as ibnashraf said

    I have another question, too.
    Another problem says, P= p ^ (q v r), Q= (p v q) ^ (p v r)
    so does the table have p, q, r columns to start off with, leaving 8 possible T/F combinations?
    yes
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    Quote Originally Posted by dude15129 View Post
    I know this is probably really easy, but I am having the hardest time understanding what I am supposed to be doing and how I am supposed to be doing it.

    It says, for each pair of propositions P & Q, state whether P & Q are logically equivalent.

    P= p^q, Q= ~p v~q

    Ok, so i don't know really where to start, I know I draw a truth table and have p q on the left side and I have TT, TF, FT, FF....then on the right side I don't know what to do.

    Any help, please?
    dude15129:

    So far you know that two statements P AND Q are logically equivalent if they have the same truth tables i.e their true-false values are the same.

    I have a question to ask you and anybody else who would care to answer.


    You know i believe the h.school theorem :

    for all x,y real Nos [absvaleu(x)<y <====> -y<x<y] i.e the two statements

    P[=absvalue(x)] and Q[= -y<x<y] are logically equivalent.

    How in this case would you test their logical equivalence???
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    Quote Originally Posted by triclino View Post
    dude15129:

    So far you know that two statements P AND Q are logically equivalent if they have the same truth tables i.e their true-false values are the same.

    I have a question to ask you and anybody else who would care to answer.


    You know i believe the h.school theorem :

    for all x,y real Nos [absvaleu(x)<y <====> -y<x<y] i.e the two statements

    P[=absvalue(x)] and Q[= -y<x<y] are logically equivalent.

    How in this case would you test their logical equivalence???
    you can't prove this from a truth table if that's what you're asking. these are not really logical statements. strictly speaking, one is the definition of the other. |x| < y means -y < x < y
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    Quote Originally Posted by Jhevon View Post
    you can't prove this from a truth table if that's what you're asking. these are not really logical statements. strictly speaking, one is the definition of the other. |x| < y means -y < x < y

    YOU mean:


    ................. absvalue(x)<y<=====> -y<x<y ,for x,y,real Nos............


    ...................is not provable??........................................ ..............
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    Quote Originally Posted by triclino View Post
    YOU mean:


    ................. absvalue(x)<y<=====> -y<x<y ,for x,y,real Nos............


    ...................is not provable??........................................ ..............
    nope, i mean it is not provable through logic, meaning through truth tables

    you can prove it based on the definition of what absolute values mean however, namely, |x| = \sqrt{x^2} = \left \{ \begin{array}{lr} x & \mbox{ if } x \ge 0 \\ & \\ -x & \mbox{ if } x < 0 \end{array} \right.
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    ..................SO when we say two statements P AND Q,are logically.....


    ....................equivalent, if they have the same true tables..........




    ..................is not generally .................................................. .........


    ...........................true?.................. .............................................

    ...................What is generally true then for logical equivalence????.


    ...The two statements in our case P[=absvalue(x)<y ] and Q[= -y<x<y] are logically equivalent
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    Quote Originally Posted by triclino View Post
    ..................SO when we say two statements P AND Q,are logically.....


    ....................equivalent, if they have the same true tables..........




    ..................is not generally .................................................. .........


    ...........................true?.................. .............................................

    ...................What is generally true then for logical equivalence????.


    ...The two statements in our case P[=absvalue(x)<y ] and Q[= -y<x<y] are logically equivalent
    the statements P and Q themselves are not logical statements that we can array in a truth table. there is simply too many things to check truth values for. you would have to do it for all real numbers x and y, which is insanity, not logic.

    the subtle difference here is between syntax and semantics. here we have a syntactic logical equivalence where you are trying to prove it to be a semantic one. it just won't work. not by truth tables. see here

    and no, logical statements and equaivalences are not always "true" in general. we can set up definitions to say whatever we want to say
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    Quote Originally Posted by Jhevon View Post
    . here we have a syntactic logical equivalence and
    There is not syntactic or semantic logical equivalence,but we can prove syntactically or semantically the logical equivalence.

    now let see you give a syntactical proof of the following equivalences:

    1) p^(qvr) <===> (p^q)v(p^r)


    2) (x)(y)[ absvalue(x)<y<====> -y<x<y], where (x),(y) means for all x,y real Nos

    since you claim to know the stuff it will be a good paradigm for those learning logic how to apply it in mathematics.

    Now the proof must be in steps and every step must be justified
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